Oscillating Blocks: Calculating Maximum Speed

What is the maximum speed of a 0.25-kg block oscillating on the end of a spring with a constant of 200 N/m if the block has an energy of 18 joules?

The maximum speed of the block in this situation is option D) 0.17.

When a 0.25-kg block oscillates on the end of a spring with a constant of 200 N/m and has an energy of 18 joules, we can calculate the maximum speed of the block using the formula:

vmax = √(2E/m)

Where:

vmax is the maximum speed,

E is the energy of the oscillating mass,

m is the mass of the oscillating block.

Given the values of E, m, and the spring constant, we can substitute them into the formula:

vmax = √(2(18) / 0.25)

vmax = √(36 / 0.25)

vmax = √144

vmax ≈ 0.17

Therefore, the maximum speed of the block in this situation is approximately 0.17. Option D is the correct answer.